Re: more on delete from join
Date: Mon, 31 Aug 2009 16:59:00 GMT
"Marshall" <marshall.spight_at_gmail.com> wrote in message news:3f70f7cf-4770-4a66-a978-33ca1dd6ced0_at_u20g2000prg.googlegroups.com...
If you want to see the value of the idea, the best way would be to compare the results given by the solving approach with any of various other approaches. The other approaches all seem to me to be ad hoc, whereas the solving approach gives a very clear theoretical foundation.
As to the particulars, the idea isn't developed yet, so the
detained comparison will have to wait.
PMFJI. Just thinking out loud.
It seems to me that an "equation" is just a special case of a relation. So the equation
x + y = 8
Is the same as a relation on x and y such that the equation is true. Of course if x and y are defined as integers, this will give a different relation than if x and y are defined as reals (or some finite subset of reals) so a system of equations is a special case of a system of relations.
x + y = 8
x - y = 2
Is a system of equations and therefore a system or relations. Now the
intersection of thses two relations is the single point, (5, 3) or, if you
prefer (x=5, y=3).
Solving the system boils down to finding the intersection of two relations. A system of equations can be overconstrained, and have no solutions (the intersection is the empty set) or underconstrained (the intersection is a set with many elements).
Now if we switch over to
A JOIN B = C and we specify that
C' = C MINUS D where D is some set of tuples to be deleted, if present
Then our job becomes to find A' and B' such that
A' JOIN B' = C' This is underconstrained, This kind of rambling, but I hope it suggests something.
. Received on Mon Aug 31 2009 - 18:59:00 CEST